# Homework Help: Longer Vector Algebra Proof

1. Oct 24, 2013

### mathnerd15

1. The problem statement, all variables and given/known data
this is an Apostol problem in chapter 12 and I guess it's a hypothetical definition of a norm of a vector
Assuming this different definition of the norm prove these statements-
$$Def. ||A||=\sum_{k=1}^{n}|a_{k}|, prove ||A||>0, if ||A||\neq0,||A||=0 if A=0, ||cA||=c||A||,triangle equality ||A+B||\leq ||A||+||B||,$$

I just looked at this- do I just expand the sums out and then express them in sigma notation?

Use this definition in V2 and prove on a figure the set of all points (x,y) of norm 1- this is just the line x+y=1?

which of the above theorems/statements would hold if we take the absolute value of the summation?

$$Def. ||A||=|\sum_{k=1}^{n}a_{k}|$$

2. Relevant equations

3. The attempt at a solution

Last edited: Oct 24, 2013
2. Oct 24, 2013

### LCKurtz

You haven't told us what $A$ is or what the $a_k$ are. If $n = 2$ is $v = \langle 1,-1\rangle \in A$? If so $v \ne 0$ but $\|v\| = 0$.

3. Oct 24, 2013

### mathnerd15

Mostly I'm asking about the triangle inequality proof, the problem just gives you a general vector A
I think the hypothetical definition is ||A||=|a1|+|a2|+|a3|...+|an|, so if ||A|| norm doesn't equal the zero vector (0,0...0) and at least one component |ak|>0, then the sum must be >0

4. Oct 24, 2013

### LCKurtz

You "think" that is the norm definition? It's your question -- is it or isn't it? I see you edited your post to change it to that. And given your last sentence, what is your question?

 Re the triangle inequality, what have you tried?

5. Oct 25, 2013

### mathnerd15

I'm just trying the TE for the first time. Apostol gives this 'incorrect' definition of the norm and asks you to prove the TE. I proved the law of cosines in norm form. here I see it is correct for a case where the signs of A and B are different

$$Def. ||A||=\sum_{k=1}^{n}|a_{k}|, triangle.inequality ||A+B||\leq ||A||+||B||. e.g.A=-2,B=1,|-1|\leq 3$$

6. Oct 25, 2013

### LCKurtz

What do you mean the signs of A and B? That doesn't make any sense. They are vectors, not scalars. You have to start by calculating A+B and its norm and compare it with the norms of A and B.

7. Oct 25, 2013

### mathnerd15

I've attached the written problems, is this how I prove the triangle equality for this definition of norm, I ended up with something that seems analogous to Cauchy-Schwarz inequality?

anyways I started to lose confidence after thinking my handwriting was too messy after I saw a video of Andrew Wiles

#### Attached Files:

• ###### Problems.pdf
File size:
664.6 KB
Views:
95
Last edited: Oct 25, 2013
8. Oct 25, 2013

### LCKurtz

You are making it way too hard. Like I said before, calculate A+B and its norm. You don't need to square anything. Just use a familiar inequality. Show your work here, not in a pdf.

9. Oct 25, 2013

### mathnerd15

$$||A+B||\leqslant ||A||+||B||\rightarrow \sum_{k=1}^{n}|ak+bk|\leq \sum_{k=1}^{n}|ak|+\sum_{k=1}^n|bk|$$
so this is trivially true by properties of absolute value? there are many more beautiful properties of absolute value functions

the last part of the solution is
|x|+|y|=1 and y=1-x, y=x+1, y=x-1, y=-x-1, and since |x|+|y|=1, then these lines become segments, 0<=x<=1 constrained and form a rectangle in V2 (similar to R2?)

Last edited: Oct 25, 2013
10. Oct 25, 2013

### LCKurtz

No, that is so wrong it leaves me speechless. One more time I will ask you:

1. Write down A = ... and B = ... and then what is A + B is.
2. Then write down its norm $\|A+B\| =$?
Then you are ready to try to get the inequality.

11. Oct 25, 2013

### mathnerd15

Sorry I'm not sure what you mean, the text gives you a hypothetical definition of norm as the sum of absolute values of |ak|
the true definition of norm as dot product gives a proof of the triangle equality from algebra or as a consequence of the Cauchy-Schwarz inequality

#### Attached Files:

• ###### IMG.pdf
File size:
264.4 KB
Views:
61
12. Oct 25, 2013

### LCKurtz

Sorry, but I will not download a pdf file for a simple problem like this. You can easily type it. But until you do 1. and 2. above, I don't see where we have anything to discuss.

The Cauchy-Schwarz inequality has nothing to do with this problem.

 I now see you have changed post #9 to make it correct, without noting that it was changed. That makes it very difficult to maintain continuity of the thread and is against forum rules.

Last edited: Oct 25, 2013
13. Oct 28, 2013

### mathnerd15

$$A=\sum_{k=1}^{n}akEk, B=\sum_{k=1}^{n}bkEk, ||A||=\sqrt{A\cdot A}. \\ ||A+B||=\sum_{k=1}^{n}[(ak+bk)^{2}]^{1/2}, \\ but here ||A||=\sum_{k=1}^{n}|ak|, ||B||=\sum_{k=1}^{n}|bk|, \\ ||A||+||B||=\sum_{k=1}^{n}(|ak|+|bk|), ||A+B||=\sum_{k=1}^{n}\|ak+bk| \\ \sum_{k=1}^{n}\|ak+bk| <=\sum_{k=1}^{n}(|ak|+|bk|)$$

I'm sorry, so all vectors in Vn can be expressed as linear combinations of the orthogonal basis vectors [Ek,En] or other arbitrarily rotated orthogonal basis vectors as well as the imaginary basis vector on the C plane? I've written what I think is the definition of norm/length and also the hypothetical one (I'm not sure if there's some significance to Apostol's hypothetical problem)

Last edited: Oct 28, 2013
14. Oct 28, 2013

### mathnerd15

actually since ||A|| is defined as the sum of absolute values then if v=(1,-1) then ||v||=2, not 0, while normally
$$||v||=2^{1/2}$$
it is really the greatest achievement of humanity! by the way what would be an example of a difficult proof to try that is more advanced or a text?

Last edited: Oct 28, 2013
15. Nov 6, 2013

### mathnerd15

am I correct in the last post?